CareerCup

o(1) space with o(n) time. Just start at the end of the sentance and scan to the beginning searching for the character. Optimal in regards to space and time.

how is that o(n) in time. You need to first go all the way to the end, which is O(n) itself. And if the last occurence of the character is the first letter itself (for example, last occurence of 'C' in "Cars are fast and expensive") , u hav to go all the way back to the beginning => O(n^2).

int lastOccurrence(char *str, char ch){

if(str==NULL)
return -1;

int pos=strlen(str)-1;
while( pos>=0 ){
if(str[pos]==ch)
break;
pos--;
}
return pos ;
}

A simple O(n) solution :

int get_position(char str[], char ch)
{
    int i, pos = 0;
    
    for(i = 0; str[i] != '\0' && i < MAX_LENGTH-1; i++)
    {
        if(str[i] == ch)
            pos = i+1;
    }
    
    return pos;
}